I am studying Griffith's Introduction to Electrodynamics and having problems to grasp few things.

Anyway, taking into account a problem of dielectric being put between parallel plate capacitor.

There is an E field of capacitor as an external field, and I know inside there is a polarization, dipole per volume P

With the story of induced charges and how dipoles are orienting, I believe I got the right in my mind. What I want to ask firstly, that field E and field P are in opposite direction? I am just asking this for example of parallel place capacitor and dielectric between them?

Now, there is a final field D, which we associate with free charges. And P is associated with the bound charges. And I should consider E fundamental and related to the total charge (free+bound).

Now: $ D=\epsilon_0E + P $

And know constantly in my head is that this should be $-P$. Like $D+p$ should equal to $E$ and also in terms of bound and free charges.

And totally I got it in my mind that $D$ and $E$ are in same direction and $P$ is opposite.

I know question might be dump, but I am losing to understand this properly, read some previous answers here and material on Internet, but still got feeling in my head "I am no exactly sure what is happening here"...

  • $\begingroup$ So all you are really asking is just "What I want to ask firstly, that field E and field P are in opposite direction? I am just asking this for example of parallel place capacitor and dielectric between them?"? Or are you asking for clarification about what these three vector fields really mean? While it is important to know where you are coming from, it seems like putting your internal struggles in the question had made it hard to know exactly what you are asking. $\endgroup$ – Aaron Stevens Sep 9 '18 at 4:41
  • $\begingroup$ Yeah, I was trying to make an general conclusion out of this example... $\endgroup$ – solidbastard Sep 9 '18 at 10:30

Just considering surface charge densities on the dielectric this is a diagram of the situation described in Griffiths.

enter image description here

The important ideal is that there is an electric field in the dielectric $\vec E_{\rm local}$ which is the sum of the electric fields due to the free charges and the bound charges. $$\vec E_{\rm local} = \vec E_{\rm free} + \vec E_{\rm bound}\Rightarrow E_{\rm local} = E_{\rm free} - E_{\rm bound}$$
with the local field being less than the field due to the free charges.

It is this local field which produces a polarization of the dielectric and the density of electric dipoles depends (linearly) on the local value of the electric field. $\vec P = \chi \epsilon_0 \vec E_{\rm local}$ where $\chi$ is the electric susceptibility of the dielectric.
Note that since $E_{\rm free} > E_{\rm bound}$ the direction of the local field $\vec E_{\rm local} $ and hence the polarization $\vec P$ will be in the same direction as $\vec E_{\rm free}$.

So defining the displacement $$\vec D = \epsilon_0 \vec E_{\rm local} + \vec P \Rightarrow D = \epsilon_0 E_{\rm local} + P$$ gives the relationship that you quoted in your question.

Furthermore if $\vec P = \chi \epsilon_0 \vec E_{\rm local}$ then $D = \epsilon_0 E_{\rm local} + \chi \epsilon_0 E_{\rm local} \Rightarrow D = \epsilon_{\rm r} \epsilon _0 E_{\rm local}$ where $\epsilon_{\rm r}(= 1+\chi)$ is the relative permittivity of the dielectric.


You define $E$ as an external field but $E$ in your equation is the final internal field. The final internal field is lower than the field expected if no dielectric was present. So in the standard case, $E$ is, say, positive, $P$ is positive, and $D$ is larger than $E$.


You appear to be getting things turned around in your mind a bit. For typical isotropic linear dielectric materials the electric susceptibility is a positive number meaning that the P and the E fields are in the same direction: $\mathbf{P}=\chi \epsilon_0 \mathbf{E}$

For the example of a parallel plate capacitor, suppose you use a fixed amount of charge on the plates. That is a free charge, so that gives a fixed amount of D. Now, if there is no dielectric then there is a certain E field leading to a certain voltage and the capacitance is the charge divided by the voltage. Next, consider what happens if there is a dielectric. The D field is the same, but now there is a P field also. Since the E and the P field are in the same direction the E field is smaller than it was without the dielectric. This means that the voltage is smaller, so dividing the same charge by a smaller voltage gives a larger capacitance.

So in the end D, E, and P are all in the same direction for linear isotropic dielectrics.

  • $\begingroup$ Thank you very much for the reply! It got some things more clear. So, I will try to think in this way now. E is always a total field. And if there is no dielectric, I got only free charges. then $D$ and are the same just the the is different for $\epsilon_0$ because $D=\epsilon_notE$ And when is there a dielectric then I got a P obviouuly, but i think I need to study more why they are in the same direction, because I thought in dielectric charges move for some small distance (positive in direction of that external field and negative in other way, making a P field in opposite of E. $\endgroup$ – solidbastard Sep 9 '18 at 12:30
  • $\begingroup$ Consider an E field vertically upwards. This means that positive charges will experience an upward force and negative charges will experience a downward force. So a dielectric will have a negative bound charge on the bottom and a positive bound charge on the top. An arrow from negative to positive points upward, so the polarization is upward, the same direction as the E field $\endgroup$ – Dale Sep 9 '18 at 12:48
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    $\begingroup$ Oh, mine. I am such an idiot I totally forgot that we defined dipole moment as a vector between two same charges, but in a direction of negative to positive one. (why is it like this, because we defined it like this? right?) Really horrible mistake, I have just made. Now it all gets fine, that polarization is in a direction of negative to positive charges and that field is in a same direction of E. And now I get that E is now reduced, then it was without dielectric; and D field is the sum of both, and represents that "move" or the "change. Thank you very much for enlighting! @Dale $\endgroup$ – solidbastard Sep 9 '18 at 13:03

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